Termination w.r.t. Q of the following Term Rewriting System could be proven:
Q restricted rewrite system:
The TRS R consists of the following rules:
*(x, *(y, z)) → *(otimes(x, y), z)
*(1, y) → y
*(+(x, y), z) → oplus(*(x, z), *(y, z))
*(x, oplus(y, z)) → oplus(*(x, y), *(x, z))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
*(x, *(y, z)) → *(otimes(x, y), z)
*(1, y) → y
*(+(x, y), z) → oplus(*(x, z), *(y, z))
*(x, oplus(y, z)) → oplus(*(x, y), *(x, z))
Q is empty.
Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:
*1(x, oplus(y, z)) → *1(x, z)
*1(+(x, y), z) → *1(x, z)
*1(x, oplus(y, z)) → *1(x, y)
*1(+(x, y), z) → *1(y, z)
*1(x, *(y, z)) → *1(otimes(x, y), z)
The TRS R consists of the following rules:
*(x, *(y, z)) → *(otimes(x, y), z)
*(1, y) → y
*(+(x, y), z) → oplus(*(x, z), *(y, z))
*(x, oplus(y, z)) → oplus(*(x, y), *(x, z))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ UsableRulesProof
Q DP problem:
The TRS P consists of the following rules:
*1(x, oplus(y, z)) → *1(x, z)
*1(+(x, y), z) → *1(x, z)
*1(x, oplus(y, z)) → *1(x, y)
*1(+(x, y), z) → *1(y, z)
*1(x, *(y, z)) → *1(otimes(x, y), z)
The TRS R consists of the following rules:
*(x, *(y, z)) → *(otimes(x, y), z)
*(1, y) → y
*(+(x, y), z) → oplus(*(x, z), *(y, z))
*(x, oplus(y, z)) → oplus(*(x, y), *(x, z))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QDPSizeChangeProof
Q DP problem:
The TRS P consists of the following rules:
*1(+(x, y), z) → *1(x, z)
*1(x, oplus(y, z)) → *1(x, z)
*1(x, oplus(y, z)) → *1(x, y)
*1(x, *(y, z)) → *1(otimes(x, y), z)
*1(+(x, y), z) → *1(y, z)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem. From the DPs we obtained the following set of size-change graphs:
- *1(x, *(y, z)) → *1(otimes(x, y), z)
The graph contains the following edges 2 > 2
- *1(x, oplus(y, z)) → *1(x, z)
The graph contains the following edges 1 >= 1, 2 > 2
- *1(x, oplus(y, z)) → *1(x, y)
The graph contains the following edges 1 >= 1, 2 > 2
- *1(+(x, y), z) → *1(x, z)
The graph contains the following edges 1 > 1, 2 >= 2
- *1(+(x, y), z) → *1(y, z)
The graph contains the following edges 1 > 1, 2 >= 2